The irreversible elementary gas-phase reaction is carried out isothermally at 305 K in a packed-bed reactor with 100 kg of catalyst. The entering pressure was 20 atm and the exit pressure is 2 atm. The feed is equal molar in A and B and the flow is in the turbulent flow regime, with FA0 10 mol/min and CA0 0.4 mol/dm3. Currently 80% conversion is achieved. What would be the conversion

0.856.

Explanation:

Lets represent the irreversible elementary gas phase equation of reaction as

A + B > C + D

We have that the percentage of conversion is 80%.

The pressure, p from the ratio of exit pressure and entering pressure is p = 2/20 = 1/10 = 0.1.

Therefore, n = 1 - p^2/ weight of the catalyst = 1 - 0.1^2/ 100 = 9.9 × 10^-3 kg cat^-.

Now, let's make use of the equation below;

J/ 1 - J = kb^2/ u [ w - nw^2/2] (1).

0.8 / 1- 0.8 = k ( 0.4)^2/ 10 [ 100 - (9.9× 10^-3 × 100^2/ 2] .

k = 4.95 dm^6/ kg.cat .mol.min

The turbulent flow= 1/2 × 9.9 × 10^-3 = 4.95 × 10^-3 kg cat^-.

Thus, making use of the equation (1) again, we have that;

{4.95 × 10^-3 × 0.4}/ 10 × [ 100 - (4.95 × 10^-3 × 100^2)] / 2 = 5.964.

Therefore, a/1 - a = 5.964.

5.964( 1 - a) = a.

5.964 - 5.964a = a.

5.964 = a + 5.964a.

5.964 = 6.964a.

a = 5.964/ 6.964 = 0.856.

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.